3.42 \(\int \frac {A+B \log (\frac {e (a+b x)}{c+d x})}{(c i+d i x)^2} \, dx\)

Optimal. Leaf size=98 \[ \frac {A (a+b x)}{i^2 (c+d x) (b c-a d)}+\frac {B (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{i^2 (c+d x) (b c-a d)}-\frac {B (a+b x)}{i^2 (c+d x) (b c-a d)} \]

[Out]

A*(b*x+a)/(-a*d+b*c)/i^2/(d*x+c)-B*(b*x+a)/(-a*d+b*c)/i^2/(d*x+c)+B*(b*x+a)*ln(e*(b*x+a)/(d*x+c))/(-a*d+b*c)/i
^2/(d*x+c)

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2525, 12, 44} \[ -\frac {B \log \left (\frac {e (a+b x)}{c+d x}\right )+A}{d i^2 (c+d x)}+\frac {b B \log (a+b x)}{d i^2 (b c-a d)}-\frac {b B \log (c+d x)}{d i^2 (b c-a d)}+\frac {B}{d i^2 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(c*i + d*i*x)^2,x]

[Out]

B/(d*i^2*(c + d*x)) + (b*B*Log[a + b*x])/(d*(b*c - a*d)*i^2) - (A + B*Log[(e*(a + b*x))/(c + d*x)])/(d*i^2*(c
+ d*x)) - (b*B*Log[c + d*x])/(d*(b*c - a*d)*i^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(42 c+42 d x)^2} \, dx &=-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{1764 d (c+d x)}+\frac {B \int \frac {b c-a d}{42 (a+b x) (c+d x)^2} \, dx}{42 d}\\ &=-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{1764 d (c+d x)}+\frac {(B (b c-a d)) \int \frac {1}{(a+b x) (c+d x)^2} \, dx}{1764 d}\\ &=-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{1764 d (c+d x)}+\frac {(B (b c-a d)) \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{1764 d}\\ &=\frac {B}{1764 d (c+d x)}+\frac {b B \log (a+b x)}{1764 d (b c-a d)}-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{1764 d (c+d x)}-\frac {b B \log (c+d x)}{1764 d (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 104, normalized size = 1.06 \[ \frac {-a A d+B (b c-a d) \log \left (\frac {e (a+b x)}{c+d x}\right )-b B (c+d x) \log (a+b x)+a B d+A b c+b B c \log (c+d x)+b B d x \log (c+d x)-b B c}{d i^2 (c+d x) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(c*i + d*i*x)^2,x]

[Out]

(A*b*c - b*B*c - a*A*d + a*B*d - b*B*(c + d*x)*Log[a + b*x] + B*(b*c - a*d)*Log[(e*(a + b*x))/(c + d*x)] + b*B
*c*Log[c + d*x] + b*B*d*x*Log[c + d*x])/(d*(-(b*c) + a*d)*i^2*(c + d*x))

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fricas [A]  time = 0.91, size = 88, normalized size = 0.90 \[ -\frac {{\left (A - B\right )} b c - {\left (A - B\right )} a d - {\left (B b d x + B a d\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{{\left (b c d^{2} - a d^{3}\right )} i^{2} x + {\left (b c^{2} d - a c d^{2}\right )} i^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^2,x, algorithm="fricas")

[Out]

-((A - B)*b*c - (A - B)*a*d - (B*b*d*x + B*a*d)*log((b*e*x + a*e)/(d*x + c)))/((b*c*d^2 - a*d^3)*i^2*x + (b*c^
2*d - a*c*d^2)*i^2)

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giac [A]  time = 0.74, size = 120, normalized size = 1.22 \[ -{\left (\frac {{\left (b x e + a e\right )} B \log \left (\frac {b x e + a e}{d x + c}\right )}{d x + c} + \frac {{\left (b x e + a e\right )} {\left (A - B\right )}}{d x + c}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^2,x, algorithm="giac")

[Out]

-((b*x*e + a*e)*B*log((b*x*e + a*e)/(d*x + c))/(d*x + c) + (b*x*e + a*e)*(A - B)/(d*x + c))*(b*c/((b*c*e - a*d
*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*(b*c - a*d)))

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maple [B]  time = 0.05, size = 515, normalized size = 5.26 \[ -\frac {B \,a^{2} d \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} \left (d x +c \right ) i^{2}}+\frac {2 B a b c \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} \left (d x +c \right ) i^{2}}-\frac {B \,b^{2} c^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} \left (d x +c \right ) d \,i^{2}}-\frac {A \,a^{2} d}{\left (a d -b c \right )^{2} \left (d x +c \right ) i^{2}}+\frac {2 A a b c}{\left (a d -b c \right )^{2} \left (d x +c \right ) i^{2}}-\frac {A \,b^{2} c^{2}}{\left (a d -b c \right )^{2} \left (d x +c \right ) d \,i^{2}}+\frac {B \,a^{2} d}{\left (a d -b c \right )^{2} \left (d x +c \right ) i^{2}}-\frac {2 B a b c}{\left (a d -b c \right )^{2} \left (d x +c \right ) i^{2}}-\frac {B a b \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} i^{2}}+\frac {B \,b^{2} c^{2}}{\left (a d -b c \right )^{2} \left (d x +c \right ) d \,i^{2}}+\frac {B \,b^{2} c \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right )^{2} d \,i^{2}}-\frac {A a b}{\left (a d -b c \right )^{2} i^{2}}+\frac {A \,b^{2} c}{\left (a d -b c \right )^{2} d \,i^{2}}+\frac {B a b}{\left (a d -b c \right )^{2} i^{2}}-\frac {B \,b^{2} c}{\left (a d -b c \right )^{2} d \,i^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln((b*x+a)/(d*x+c)*e)+A)/(d*i*x+c*i)^2,x)

[Out]

-1/(a*d-b*c)^2/i^2*A*b*a+1/d/(a*d-b*c)^2/i^2*A*b^2*c-d/(a*d-b*c)^2/i^2*A/(d*x+c)*a^2+2/(a*d-b*c)^2/i^2*A/(d*x+
c)*a*b*c-1/d/(a*d-b*c)^2/i^2*A/(d*x+c)*b^2*c^2-1/(a*d-b*c)^2/i^2*B*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*b*a+1/d/(a*
d-b*c)^2/i^2*B*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*b^2*c-d/(a*d-b*c)^2/i^2*B*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)/(d*x+
c)*a^2+2/(a*d-b*c)^2/i^2*B*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)/(d*x+c)*a*b*c-1/d/(a*d-b*c)^2/i^2*B*ln(b/d*e+(a*d-b
*c)/(d*x+c)/d*e)/(d*x+c)*b^2*c^2+d/(a*d-b*c)^2/i^2*B/(d*x+c)*a^2-2/(a*d-b*c)^2/i^2*B/(d*x+c)*a*b*c+1/d/(a*d-b*
c)^2/i^2*B/(d*x+c)*b^2*c^2+1/(a*d-b*c)^2/i^2*B*b*a-1/d/(a*d-b*c)^2/i^2*B*b^2*c

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maxima [A]  time = 1.10, size = 134, normalized size = 1.37 \[ -B {\left (\frac {\log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right )}{d^{2} i^{2} x + c d i^{2}} - \frac {1}{d^{2} i^{2} x + c d i^{2}} - \frac {b \log \left (b x + a\right )}{{\left (b c d - a d^{2}\right )} i^{2}} + \frac {b \log \left (d x + c\right )}{{\left (b c d - a d^{2}\right )} i^{2}}\right )} - \frac {A}{d^{2} i^{2} x + c d i^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^2,x, algorithm="maxima")

[Out]

-B*(log(b*e*x/(d*x + c) + a*e/(d*x + c))/(d^2*i^2*x + c*d*i^2) - 1/(d^2*i^2*x + c*d*i^2) - b*log(b*x + a)/((b*
c*d - a*d^2)*i^2) + b*log(d*x + c)/((b*c*d - a*d^2)*i^2)) - A/(d^2*i^2*x + c*d*i^2)

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mupad [B]  time = 4.89, size = 106, normalized size = 1.08 \[ -\frac {A-B}{x\,d^2\,i^2+c\,d\,i^2}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}{d^2\,i^2\,\left (x+\frac {c}{d}\right )}+\frac {B\,b\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d\,i^2\,\left (a\,d-b\,c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log((e*(a + b*x))/(c + d*x)))/(c*i + d*i*x)^2,x)

[Out]

(B*b*atan((b*c*2i + b*d*x*2i)/(a*d - b*c) + 1i)*2i)/(d*i^2*(a*d - b*c)) - (B*log((e*(a + b*x))/(c + d*x)))/(d^
2*i^2*(x + c/d)) - (A - B)/(d^2*i^2*x + c*d*i^2)

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sympy [B]  time = 1.52, size = 231, normalized size = 2.36 \[ \frac {B b \log {\left (x + \frac {- \frac {B a^{2} b d^{2}}{a d - b c} + \frac {2 B a b^{2} c d}{a d - b c} + B a b d - \frac {B b^{3} c^{2}}{a d - b c} + B b^{2} c}{2 B b^{2} d} \right )}}{d i^{2} \left (a d - b c\right )} - \frac {B b \log {\left (x + \frac {\frac {B a^{2} b d^{2}}{a d - b c} - \frac {2 B a b^{2} c d}{a d - b c} + B a b d + \frac {B b^{3} c^{2}}{a d - b c} + B b^{2} c}{2 B b^{2} d} \right )}}{d i^{2} \left (a d - b c\right )} - \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}}{c d i^{2} + d^{2} i^{2} x} + \frac {- A + B}{c d i^{2} + d^{2} i^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)**2,x)

[Out]

B*b*log(x + (-B*a**2*b*d**2/(a*d - b*c) + 2*B*a*b**2*c*d/(a*d - b*c) + B*a*b*d - B*b**3*c**2/(a*d - b*c) + B*b
**2*c)/(2*B*b**2*d))/(d*i**2*(a*d - b*c)) - B*b*log(x + (B*a**2*b*d**2/(a*d - b*c) - 2*B*a*b**2*c*d/(a*d - b*c
) + B*a*b*d + B*b**3*c**2/(a*d - b*c) + B*b**2*c)/(2*B*b**2*d))/(d*i**2*(a*d - b*c)) - B*log(e*(a + b*x)/(c +
d*x))/(c*d*i**2 + d**2*i**2*x) + (-A + B)/(c*d*i**2 + d**2*i**2*x)

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